Question

http://www.ms.uky.edu/~ma123/
Go to Old Exams on the left
Then click Exam 3 with answers from the fall of 2012
Then go to problem #5

Answer 1

It gives you the answer but I need an explanation.

Answer 2

You would use the product rule to find the derivative. The inflection point is the second derivative set equal to 0.

Answer 3

can you solve it out for me? sorry if im being a pain. :P

Answer 4

It looks like they actually give you the second derivative so you don't have to find it, just set it equal to zero.

Answer 5

x-3=0 so x=3

Answer 6

yeah i saw. what happens to the denominator?

Answer 7

Multiply it out and multiplying anything *0 =0, so x-3=0

Answer 8

what about #6?

Answer 9

We know that it is a parabola because it is x^2 meaning that it will only have one max or min so you automatically know that it goes to either Inf or -Inf and concave up is just where the critical value in the second derivative equation is positive.

Answer 10

@KingGeorge

Answer 11

#8?

Answer 12

We want to maximize \(A=x\cdot y\) of the rectangle where \(A\) is the area. However, since we have the relation \(y=1/(x^2+4)\), we can substitute, and get the relation\[A={x \over x^2+4}\]Now we just need to maximize it.

Answer 13

To do this, we need to take the derivative, so\[A' = {4-x^2 \over (4+x^2)^2}\]Which has zeroes at \(x=\pm 2\). Since length is positive, we know \(x=2\). Using that, plug it into our formula for \(A\), so we get \[A= {2 \over 2^2+4}={2 \over 8}={1 \over 4}\]

Answer 14

why is #13 A NOT C?

Answer 15

@amistre64